I am a five digit even number.My hundred digit is 1 more than my tens digit .My thousands digits is dibisible by 5 and is greater than 2 .my ten thousands digit is the same as my ones digits .What number am l?
Something is wrong here: "My thousands digits is dibisible by 5 and is greater than 2 ." If the digit is divisible by 5, it is 5. Saying it is greater than 2 makes no sense. I suspect you are missing something.
To find the number that satisfies the given conditions, we can break down the information step by step:
1. The number is a five-digit even number. This means that the ones digit must be an even number (0, 2, 4, 6, or 8).
2. The hundred digit is 1 more than the tens digit. Let's use a variable to represent the tens digit. Let's say the tens digit is 'x', which means the hundred digit would be 'x + 1'.
3. The thousands digit is divisible by 5 and is greater than 2. This means the thousands digit must be 5 or 10.
4. The ten-thousands digit is the same as the ones digit. We already know the ones digit is an even number, so the ten-thousands digit must also be an even number.
Putting all this information together, we can try different possibilities until we find a number that satisfies all conditions:
Case 1: If the ten-thousands digit is 0, the ones digit must also be 0. But in this case, the hundred digit would be 1, breaking the condition that the thousands digit is divisible by 5 and greater than 2. Therefore, this case is not valid.
Case 2: If the ten-thousands digit is 2, the ones digit must also be 2. Now, the number involves '2x1'. We have two remaining digits: one for the thousands digit and one for the tens digit.
We have two possible cases for the thousands digit:
- If the thousands digit is 10, then the remaining digit for the tens place would be 3. So, the number would be 20321, but it is not an even number since the ones digit is odd.
- If the thousands digit is 5, then the remaining digit for the tens place would be 4. So, the number would be 54221, which is an even number.
Therefore, the number that satisfies all the given conditions is 54221.